Question 193069
In triangle ABC, a=6, b=7, and cos C=1/4

A.) Find the exact area of the triangle. 
B.) Find C
C.) Find sin A in simplest radical form.

The part I'm having trouble with it what does cos C=1/4 mean?

{{{Cos(C)=1/4}}} means that the cosine of C is {{{1/4}}}

The formula for the area is

{{{A = (ab*Sin(C))/2}}}

So first we must find {{{sin(C)}}} from {{{cos(C)}}}

To do this we make use of the identity {{{Sin^2alpha+Cos^2alpha=1}}}

{{{Sin^2C+Cos^2C=1}}}

{{{Sin^2C=1-Cos^2C}}}

Since we know that {{{Cos(C)=1/4}}}

{{{Sin^2C=1-(1/4)^2}}}
{{{Sin^2C=1-1/16}}}
{{{Sin^2C=16/16-1/16}}}
{{{Sin^2C=15/16}}}
{{{Sin(C)=sqrt(15/16)}}}
{{{Sin(C)=sqrt(15)/4}}}

Now we can substitute in the area formula:

{{{A = (a*b*Sin(C))/2}}}

{{{A = (6*7*(sqrt(15)/4))/2}}}

{{{A = (6*7*(sqrt(15)/4))*(2/1)}}}

{{{A=(6*7*2/4)sqrt(15)}}}

{{{A=21sqrt(15)}}}

Edwin